integral of exponential of Brownian motion I am currently reading a proof that uses the following fact without proof:

If $B$ is a scalar standard Brownian motion, then $\int_0^\infty e^{B_s} \,ds = + \infty$ a.s..

How can we justify this fact? I don't see how this follows from any property of Brownian motion.
 A: Let $A$ be the event that $\int_0^\infty e^{B_t} \,dt < +\infty$.  By the Kolomorgov 0-1 law, $P(A) = 0$ or $1$.
Now let $B$ be the event that $\int_0^\infty e^{-B_t}\,dt < +\infty$.  By symmetry, $P(A) = P(B)$.  Moreover, on $A \cap B$ we have $$\int_0^\infty (e^{B_t} + e^{-B_t}) \,dt < +\infty$$
which is absurd since $e^x + e^{-x} \ge 1$ for all $x$.  So $A \cap B = \emptyset$.  This makes it impossible that $P(A) = P(B) = 1$, so we must have $P(A) = 0$.
Another approach (more involved) is to use recurrence and the strong Markov property to show that almost surely, there are infinitely many disjoint intervals of length 1 on which $B_t$ stays above $-1$.
A: (Not a proof, but here's some intuition at least) 
Standard Brownian motion has mean 0 and variance t for $0 \le t \lt \infty$
Thus, on average the integral becomes...
$$\int_0^{\infty} e^0 \ ds=\int_0^{\infty} 1 \ ds$$
Which clearly diverges. What you should aim for proving is that Brownian motion crosses between negative and positive values in such a fashion that as the limit of the integral increases to infinity, the number of reversals also increases to infinity with a mean period between oscillations that isn't 0.
